Minimum Factorization

Problem Statement

The task revolves around finding the smallest possible positive integer ( x ) where the product of its individual digits equals a given integer ( num ). For example, for ( x ) being 68, the product of 6 and 8 results in 48. This problem comes with several considerations:

1. The resulting integer ( x ) must be the smallest in value that meets the product requirement.
2. If no such integer ( x ) can satisfy the product condition or it exceeds the 32-bit signed integer range, the function must return 0.

This requires a systematic approach to break down the given number and test various combinations of factors efficiently while considering performance and feasibility within the given constraints.

Examples

Example 1

Input:

num = 48

Output:

68

Example 2

Input:

num = 15

Output:

35

Constraints

• 1 <= num <= 231 - 1

Approach and Intuition

The problem can be approached by trying to factorize the number ( num ) systematically using digits from 1 to 9 (since any digit greater than 9 would not be applicable in forming a direct multiplier).

1. Start with the largest possible digit, 9, and work downwards to see if ( num ) is divisible by these digits.
2. For each digit that divides ( num ) exactly, reduce ( num ) by repeatedly dividing it by the digit until it no longer can be divided completely by that digit. This maximizes the use of higher digits, making ( x ) smaller.
3. The digits collected through these divisions will form the integer ( x ). However, since the divisions are performed with the largest factors first, the collected digits will be in reverse order.
4. If after processing all possible digits from 9 to 1, ( num ) reduces to 1, this suggests the digits collected can form the required product. If ( num ) is greater than 1, it indicates that ( num ) includes a prime factor greater than 9, mandating a return of 0, as no collection of base-10 digits could achieve that product.
5. Finally, compile the collected digits in the correct order and verify whether the resultant integer fits within a 32-bit signed integer. If it exceeds, return 0.

Considering an example: for ( num = 48 ):

• Starting with 9, ( num ) is not divisible by 9, next check 8, and 48/8 = 6.
• 6 itself can be divided by digits down to 1 but favoring the larger digits gives us 6.
• Arranging our results gives us 68.

In cases where ( num ) is impossible to break down strictly into digits (such as when ( num ) contains a prime factor greater than 9), or the resultant number exceeds the 32-bit limit, the solution defaults to 0. The systematic exploration of digit options from largest to smallest ensures the smallest possible resultant ( x ). This approach adequately simplifies the solution while adhering to the constraints provided.

Solutions

public class Solution {
    public int minimumFactorization(int number) {
        if (number < 2)
            return number;
        long result = 0, multiplier = 1;
        for (int i = 9; i >= 2; i--) {
            while (number % i == 0) {
                number /= i;
                result = multiplier * i + result;
                multiplier *= 10;
            }
        }
        return number < 2 && result <= Integer.MAX_VALUE ? (int)result : 0;
    }
}